Question

Continuity on a closed Interval In Exercises 31-34, discuss the continuity of the function on the closed interval. $$ f(x)=\left\\{\begin{array}{ll}{3-x,} & {x \leq 0} \\ {3+\frac{1}{2} x,} & {x>0}\end{array} \quad[-1,4]\right. $$

Short answer

The function \(f(x)\) given by the equation,\[f(x)=\begin{array}{ll}{3-x,} & {x \leq 0} \ {3+\frac{1}{2} x,} & {x>0}\end{array}\], is continuous on the closed interval \([-1,4]\)

Step-by-step solution

Identify the Function

The function \(f(x)\) is given as: \[f(x)=\begin{array}{ll}{3-x,} & {x \leq 0} \ {3+\frac{1}{2} x,} & {x>0}\end{array}\]. It is defined differently for \(x \leq 0\) and \(x > 0\)

Substitute the Range Endpoints

Evaluate the function at the endpoints of the specified range [-1,4]. For \(x=-1\) and \(x=0\), we need to use the first line of the piecewise function, while for \(x=4\), the second line. When \(x=-1\), the function is equal to \(3-(-1)=4\). When \(x=0\), the function is \(3-0=3\). When \(x=4\), the function is \(3+1/2*4=5\).

Determine Continuity at the Transition

Now we must evaluate the function at the point where the two parts of the function meet, which is at \(x = 0\). We have already found that \(f(x=0) = 3\) using the first piece. We also want to check that \(x=0\) for the second piece will also be \(3\). Using \(x=0\), the function would be equal to \(3+1/2*0=3\). The function is equal at this point, therefore, it is continuous at \(x=0\) because the left-hand limit and the right-hand limit yield the same result.

Conclude the Continuity

Since evaluated 'f' at any point within the range [-1,4] including the transition point yields well-defined and equal results, we can conclude that \(f(x)\) is continuous over the interval \([-1,4]\)

Key concepts

Piecewise Functions

Piecewise functions are mathematical constructs where a single function is defined by different expressions over various parts of its domain. They are particularly useful for describing situations like a cooling rate or tax brackets, where the rule that defines the function changes at certain points. These functions can be written in a format that clearly indicates the different expressions and the conditions under which each one applies. For the function given in the exercise

• When \(x \leq 0\), the function is described by the expression \(3-x\).
• When \(x > 0\), it switches to \(3+\frac{1}{2}x\).

This step-wise definition makes it easy to select the correct rule for a given input value. In evaluating continuity, it’s essential to understand each segment of a piecewise function, as what’s being used can change at the boundaries or transition points of the function.

Closed Interval

A closed interval in mathematics refers to an interval that includes all its endpoints. When examining functions on a closed interval, like defined by this ensures that the function needs to be evaluated not only throughout each segment but also right at the boundaries or edges of the interval. In our exercise, the closed interval [-1, 4] includes both endpoints -1 and 4. To check for continuity over this interval, you must ensure the function is defined at both ends and behaves continuously across its entire range. Having endpoints included helps in covering every possible input value within the interval, ensuring comprehensive analysis during evaluations.

Endpoint Evaluation

Evaluating a function at its endpoints is crucial when assessing its continuity over a closed interval. Endpoints are where the behavior of the function might change or where its definition could be ambiguous. For instance, in the function defined above, endpoints like \(-1\) and \(4\) need specific examination. By substituting \(-1\), the calculation results in \(f(-1) = 4\) since it falls under the expression \(3-x\). Similarly, substituting \(4\) under \(3+\frac{1}{2}x\) yields an output of \(5\).An accurate endpoint evaluation is significant because it guarantees the values are consistent with the function's equations. It is also a necessary step in determining if the function maintains its continuity throughout the given range.

Left-Hand and Right-Hand Limits

Left-hand and right-hand limits help in understanding how a function behaves as it approaches a certain point from either direction. This is especially important for piecewise functions at their transition points to verify continuity. For instance, suppose you have a function defined piecewise, the left-hand limit at \(x=0\) can be calculated using the expression that holds for \(x \leq 0\):\[ \lim_{{x \to 0^-}} f(x) = f(0) = 3 \]The right-hand limit, on the other hand, approaches from \(x > 0\):\[ \lim_{{x \to 0^+}} f(x) = 3 + \frac{1}{2} \times 0 = 3 \]When both limits converge to the same value at the transition point, the function is continuous at that point. In this exercise, consistency between the left and right-hand limits at \(x=0\) confirms that there is no discontinuity, ensuring smoothness of the function across its entire domain. This bridging of the two segments at \(0\) solidifies the function's overall continuity.